The equation is Czech-Slovak Match 1996 and solution is very easy. Solution. The equation can be rewritten as $p^x=y^p+1.$ Now $y=1$ leads to $(p,x)= (2,1)$ and $(y,p)=(2,3)$ leads to $x=2.$ For $y>1$ and $p\not= 3,$ by $Zsigmondy's$ theorem,$y^p+1$ has a prime factor that does not divide $y+1.$ Since $y^p+1$ is divisible by $y+1,$ it follows that $y^p+1$ has at least two prime factors, a contradiction.

also here a correction, it is from Czech–Slovak Match 1995, P6 according to Imomath

Ferid.---. wrote: The equation is Czech-Slovak Match 1996 and solution is very easy. Solution. The equation can be rewritten as $p^x=y^p+1.$ Now $y=1$ leads to $(p,x)= (2,1)$ and $(y,p)=(2,3)$ leads to $x=2.$ For $y>1$ and $p\not= 3,$ by $Zsigmondy's$ theorem,$y^p+1$ has a prime factor that does not divide $y+1.$ Since $y^p+1$ is divisible by $y+1,$ it follows that $y^p+1$ has at least two prime factors, a contradiction. I try to make a solution without using Zsigmondy and it results that LTE works really well. But obviously is not so short as this.However I'll post it anyway: $p^x=y^p+1.$ first we handle the case $p=2$ and we get: $2^x=y^2+1$ , obviously $y$ is odd that means that $y^{2} \equiv 1 (\text{mod} 4) $ $$ \implies y^2+1\equiv 2(\text{mod} 4)$$that means that $x<2$ $ \implies x=1$ $ \implies y=1$ then if we try the case $p>2$ we use LTE: I wrote it in the file below is really extensive.

Isn't $(x,y,p)=(0,0,p)$ a solution as the question says $x,y$ non-negetive? In that case I think the solution is $ (x,y,p) = (0,0,p) (1,1,2) (2,2,3) $